Thread: Cube number

I'm wondering if there is any way to prove that of the numbers in the form 1.....1 (i.e. all digits are 1), 1 is the only cube number.

Originally Posted by BenWong

I'm wondering if there is any way to prove that of the numbers in the form 1.....1 (i.e. all digits are 1), 1 is the only cube number.

i was able to narrow it down a bit. i could prove that the only candidates for being a cube are the numbers of the form .
i can post the proof of the above if you are familiar with modular arithmetic and fermat's little theorem.

That'd be great! I took abstract algebra a while ago, but I should be able to follow.

Originally Posted by BenWong

That'd be great! I took abstract algebra a while ago, but I should be able to follow.

first we use the fact that .
proof: using fermat's li'l theorem so either or . the latter implies .

so if has to be a cube then the inly admissible values of would be the ones satisfying . there are only trials needed to know which are those. the possible values of are . that's it.
in my last post i missed the numbers of the type .

Originally Posted by BenWong

I'm wondering if there is any way to prove that of the numbers in the form 1.....1 (i.e. all digits are 1), 1 is the only cube number.

According to this book by Paulo Ribenboim, this result was proved by Andrzej Rotkiewicz in 1987. But I do not have a reference for the proof.

So is the next step to show that a number 1....1 can never be written in those three forms. How do you show that?

Thanks for the help.

Originally Posted by BenWong

So is the next step to show that a number 1....1 can never be written in those three forms. How do you show that?

Thanks for the help.

well i could not go further as i had already said "i could only narrow it down a bit".